One observer estimates the angle of elevation to the basket of a hot air balloon to be 60°, while another observer 100 m away estimates the angle of…

Let B is the basket of hot air balloon

\[\begin{aligned} & \tan 60^{\circ}=\sqrt{3}=\frac{h}{x} \Rightarrow h=x \sqrt{3} \\ & \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{h}{x+100} \Rightarrow x=h \sqrt{3}-100 \end{aligned} \]

using (i) and (ii)

# ANOTHER SOLUTION AS PER BELOW FIGURE IS ALSO POSSIBLE

Let \(\displaystyle B\) is the basket of hot air balloon. \(\displaystyle D\) and \(\displaystyle C\) be the positions of the first and second observer's respectively.

\[\begin{aligned} & \tan 60^{\circ}=\sqrt{3}=\frac{h}{x} \Rightarrow h=x \sqrt{3} \\ & \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{h}{100-x} \Rightarrow \sqrt{3} h=100-x \end{aligned} \]

using (i) and (ii)

\[\begin{aligned} & h=\sqrt{3}(100-\sqrt{3} h) \Rightarrow h=25 \sqrt{3} \mathrm{~m} \\ & \text { (b) } \sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{h}{B D} \Rightarrow B D=50 \mathrm{~m} \\ & \text { (c) } \mathrm{x}=\frac{h}{\sqrt{3}}=25 \mathrm{~m} \\ & \Rightarrow \mathrm{AC}=100-\mathrm{x}=75 \mathrm{~m} \end{aligned} \]